An object with a mass of 2 kg, temperature of 4 ^oC, and a specific heat of 19 J/(kg*K) is dropped into a container with 24 L of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Narad T.
Dec 12, 2017

The water does not evaporate and the change in temperature is =0.002^@C

Explanation:

The heat is transferred from the hot object to the cold water.

Let T= final temperature of the object and the water

For the cold water, Delta T_w=T-0=T

For the object DeltaT_o=4-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

The specific heat of water is C_w=4.186kJkg^-1K^-1

The specific heat of the object is C_o=0.019kJkg^-1K^-1

The mass of the object is m_0=2kg

The mass of the water is m_w=24kg

2*0.019*(4-T)=24*4.186*T

4-T=(24*4.186)/(2*0.019)*T

4-T=2643.8T

2644.8T=4

T=4/2644.8=0.002^@C

As T<100^@C, the water does not evaporate

You would expect this result, as the object is not very hot and the mass of the object and its specfic heat are small compared to the mass of water and the specific heat of water.

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